Ian: New page: We first calculate the particle´s acceleration using Newton´s Second Law $F=ma$ . $\begin{align} & 15=6a \\ & \text{giving} \\ & a=2\textrm{.}5\ \text{m}{{\text{s}}^{-...$

Ian — Fri, 21 May 2010 15:18:22 GMT

New page: We first calculate the particle´s acceleration using Newton´s Second Law <math>F=ma</math>. <math>\begin{align} & 15=6a \\ & \text{giving} \\ & a=2\textrm{.}5\ \text{m}{{\text{s}}^{-...

New page

We first calculate the particle´s acceleration using Newton´s Second Law <math>F=ma</math>.

<math>\begin{align}
& 15=6a \\
& \text{giving} \\
& a=2\textrm{.}5\ \text{m}{{\text{s}}^{-2}} \\
\end{align}</math>

Now using the kinematic equation (see section 6),

<math>v=u+at\ </math> gives

<math>\begin{align}
& 10=0+2\textrm{.}5\times t \\
& \\
& t=4\ \text{s} \\
\end{align}</math>

Solution 10.4 - Revision history

Ian: New page: We first calculate the particle´s acceleration using Newton´s Second Law F=ma. \begin{align} & 15=6a \\ & \text{giving} \\ & a=2\textrm{.}5\ \text{m}{{\text{s}}^{-...

Ian: New page: We first calculate the particle´s acceleration using Newton´s Second Law $F=ma$ . $\begin{align} & 15=6a \\ & \text{giving} \\ & a=2\textrm{.}5\ \text{m}{{\text{s}}^{-...$